\(\left(sinx-cosx\right)\left(sin^2x+cos^2x+sinx.cosx\right)-1=3sin2x\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1+sinx.cosx\right)-6sinx.cosx-1=0\)
Đặt \(sinx-cosx=t\) (\(\left|t\right|\le\sqrt{2}\))
\(\Rightarrow t^2=sin^2x+cos^2x-2sinx.cosx\Rightarrow sinx.cosx=\frac{1-t^2}{2}\)
Phương trình trở thành:
\(t\left(1+\frac{1-t^2}{2}\right)-3\left(1-t^2\right)-1=0\)
\(\Leftrightarrow t^3-6t^2-3t+8=0\)
\(\Leftrightarrow\left(t-1\right)\left(t^2-5t-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=1\\t=\frac{5+\sqrt{57}}{2}>\sqrt{2}\left(l\right)\\t=\frac{5-\sqrt{57}}{2}< -\sqrt{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow sinx-cosx=1\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
